本文实例讲述了PHP查询附近的人及其距离的实现方法。分享给大家供大家参考,具体如下:
<?php //获取该点周围的4个点 $distance = 1;//范围(单位千米) $lat = 113.873643; $lng = 22.573969; define('EARTH_RADIUS', 6371);//地球半径,平均半径为6371km $dlng = 2 * asin(sin($distance / (2 * EARTH_RADIUS)) / cos(deg2rad($lat))); $dlng = rad2deg($dlng); $dlat = $distance/EARTH_RADIUS; $dlat = rad2deg($dlat); $squares = array('left-top'=>array('lat'=>$lat + $dlat,'lng'=>$lng-$dlng), 'right-top'=>array('lat'=>$lat + $dlat, 'lng'=>$lng + $dlng), 'left-bottom'=>array('lat'=>$lat - $dlat, 'lng'=>$lng - $dlng), 'right-bottom'=>array('lat'=>$lat - $dlat, 'lng'=>$lng + $dlng) ); print_r($squares['left-top']['lat']); //从数库查询匹配的记录 $info_sql = "select * from `A` where lat<>0 and lat>{$squares['right-bottom']['lat']} and lat<{$squares['left-top']['lat']} and lng>{$squares['left-top']['lng']} and lng<{$squares['right-bottom']['lng']} "; //获取两点之间的距离 function getDistanceBetweenPointsNew($latitude1, $longitude1, $latitude2, $longitude2) { $theta = $longitude1 - $longitude2; $miles = (sin(deg2rad($latitude1)) * sin(deg2rad($latitude2))) + (cos(deg2rad($latitude1)) * cos(deg2rad($latitude2)) * cos(deg2rad($theta))); $miles = acos($miles); $miles = rad2deg($miles); $miles = $miles * 60 * 1.1515; $feet = $miles * 5280; $yards = $feet / 3; $kilometers = $miles * 1.609344; $meters = $kilometers * 1000; return compact('miles','feet','yards','kilometers','meters'); } $point1 = array('lat' => 40.770623, 'long' => -73.964367); $point2 = array('lat' => 40.758224, 'long' => -73.917404); $distance = getDistanceBetweenPointsNew($point1['lat'], $point1['long'], $point2['lat'], $point2['long']); foreach ($distance as $unit => $value) { echo $unit.': '.number_format($value,4).'<br />'; } ?>
希望本文所述对大家PHP程序设计有所帮助。
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